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933. Increasing Order Search Tree

Stack Tree Depth-First Search Binary Search Tree Binary Tree

Problem - Increasing Order Search Tree

Easy

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

 

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

 

Constraints:

  • The number of nodes in the given tree will be in the range [1, 100].
  • 0 <= Node.val <= 1000

Solutions

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def increasingBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        dummy = last = TreeNode(right=root)
        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            nonlocal last
            dfs(root.left)
            last.right = root
            root.left = None
            last = root
            dfs(root.right)

        dfs(root)
        return dummy.right

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 17.6 MB (89.09%)